Question 1136613
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<pre>
When the friend left the beach, Lucy was  (0.5 hours) * 50 mph = 25 mph ahead.


But the friend moves faster, decreasing the distance between them by 5 = 55 - 50 miles per hour.


So, the friend will catch Lucy in  {{{25/5}}} = 5 hours.
</pre>


<U>Algebra solution</U>


<pre>
Algebra solution is this equation


    50*(t+0.5) = 55*t,


saying that both of them cover the same distance from the start to the catching point.


In this equation  " t " is the time starting from the moment, when the friend began his move.


Simplify and solve it


    50t + 25 = 55t

    25 = 55t - 50t

    25 = 5t

    t = {{{25/5}}} = 5 hours.


The same answer as in the very first solution.
</pre>

Solved.


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See introductory lessons on Travel and Distance problems

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems.lesson>Travel and Distance problems</A>  

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems-for-two-bodies-moving-toward-each-other.lesson>Travel and Distance problems for two bodies moving in opposite directions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Typical-catching-up-Travel-and-Distance-problems.lesson>Travel and Distance problems for two bodies moving in the same direction (catching up)</A>

in this site.


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Read them and learn once and for all from these lessons on how to solve simple Travel and Distance problems.


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Ignore the post by @josgarithmetic, since it is totally nonsensical.